Question

$\frac{\sin^3 \, 3A}{\sin^2 A} - \frac{\cos^2 3A}{\cos^2 A} = $

• A. $\frac{1}{8 \cos \, 2A}$
• B. 8 cos 2A
• C. cos 2A
• D. none of these

Answer 1

Answer: (B)

$\frac{\sin^{2 }3A}{\sin^{2}A} - \frac{\cos^{2} 3A}{\cos^{2}A} $
$= \left(\frac{3 \sin A - 4 \sin^{3} A}{\sin A}\right)^{2} - \left(\frac{4\cos^{3} A - 3 \cos A}{\cos A}\right)^{2}$
$ = \left(3-4 \sin^{2} A\right)^{2} - \left(4 \cos^{2} A - 3\right)^{2} $
$= 16 \sin^{4} A - 16 \cos^{4} A - 24 \left(\sin^{2} A - \cos^{2} A\right)$
$= 16\left(\sin^{2} A - \cos^{2}A\right) \left(\sin^{2} A +\cos^{2} A\right) - 24 \left(\sin^{2} A - \cos^{2} A\right) $
$= - 8 \left(\sin^{2} A - \cos^{2} A\right) = 8 \cos2A$